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Pragma · 07-11-2007, 05:46 PM

That's because you're referencing it wrong.

Code:

printf("byteBuffer: %X \n", byteBuffer[0] );

Try that instead - I think what's happening instead is that you're printing a the address of the char* byteBuffer pointer.

07-11-2007, 05:46 PM	#4 (permalink)
Pragma I am Winter Born Location: Alexandria, VA	That's because you're referencing it wrong. Code: printf("byteBuffer: %X \n", byteBuffer[0] ); Try that instead - I think what's happening instead is that you're printing a the address of the char* byteBuffer pointer. __________________ Eat antimatter, Posleen-boy!